Question: Factor completely. $100 -140x +49x^2=$
Explanation: Both $100$ and $49x^2$ are perfect squares, since $100=({10})^2$ and $49x^2=({7x})^2$. Additionally, $140x$ is twice the product of the roots of $100$ and $49x^2$, since $140x=2({10})({7x})$. $100x^2-140x+49 = ({10})^2 - 2({10})({7x})+({7x})^2$ So we can use the square of a difference pattern to factor: ${a}^2 - 2( a)( b)+ {b}^2 =({a} - {b})^2$ In this case, ${a}={10}$ and ${b}={7x}$ : $ ({10})^2 - 2({10})({7x})+({7x})^2 =({10} - {7x})^2$ In conclusion, $100 -140x +49x^2=(10 -7x)^2$ Remember that you can always check your factorization by expanding it.